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Consider triangles ABC, EFG that have
AB=EF, ^CAB= ^GEF and ^ABC = ^EFG
Move ∇ABC so that AB lies on EF with B positioned upon F
A will lie upon E as AB=EF
AC will lie along EG as ^CAB=^GEF
BC will lie along FG as ^ABC=^EFG
C will coincide with G (straight lines can only intersect at a single point)
∴ ∇ABC and ∇EFG are identical
Created 1st Aug 2008
The above "proof" ignores the possibilty of one triangle being a mirror image of the other. In Fig. 2 ∇ABC and ∇PQR can be overlaid as described previously, but ∇P'Q'R' cannot even though it has the equivalent side and its two ajacent angles equal to ∇ABC, i.e.
BC=Q'R', ^ABC=^P'R'Q' and ^ACB=^P'R'Q'.
However if ∇P'R'Q' is moved so that point Q' overlays point B and R'Q'BC is a straight line.
YX is perpendicular to RC.
If ∇P'R'Q' is rotated about YX, R' will overlay C as R'Q'=BC and R'C is a straight line.
P'Q' will Align with AB as ^P'Q'R'=^ABC
P'R' will Align with AC as ^P'R'Q'=^ACB
So P' must overlay A
∴∇P'Q'R'=∇ABC
These proofs do not enthuse me and lack elegance and no doubt are not mathematically rigorous but satisfy me enough so that I can start using equal triangles to explore much more interesting geometry.
Updated November 2019
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